Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 108 km/h brakes with a constant deceleration of 8 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 7 m/s from the edge of a roof 25 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
An 18 kg block slides down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 7.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 45° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is thrown from ground level at 35 m/s and 70° above the horizontal. It lands on the flat roof of a building 3 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A small ball on a string of length 0.8 m moves in a horizontal circle so that the string makes a constant angle of 30° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the speed v of the ball.
A 0.9 kg ball on a string 0.6 m long is swung in a vertical circle. At the lowest point it moves at 4 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A block slides from rest down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 7 m along the slope.
A 2 kg block rests on a rough horizontal table (μ = 0.4). A light string runs from it over a smooth pulley at the edge of the table to a hanging 1 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 3 m.
A 1 kg glider moving at 9 m/s makes a perfectly elastic head-on collision with a 2 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 1700 kg car runs into a parked 1400 kg car. The wrecks lock together and skid 12 m before stopping. The coefficient of kinetic friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A uniform slender rod of mass 16 kg and length 2.3 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find its angular acceleration α just after release.
A hollow sphere rolls without slipping down a slope inclined at 25° to the horizontal. Use g = 9.81 m/s². Find the smallest coefficient of static friction μs that prevents it from slipping.
A belt runs on a 150 mm driving pulley at 720 rpm. The tension is 600 N on the tight side and 150 N on the slack side. Find the power P transmitted, in kW.
A 7.5 kW motor at 2900 rpm drives gear A (12 teeth), which meshes with gear B (36 teeth). Gear C (20 teeth) on the same shaft as B drives gear D (24 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
An astronaut on another planet finds that a simple pendulum 0.5 m long has a period of 2.31 s for small swings. Find the gravitational acceleration g on that planet.
A 2.1 kg mass on a spring with k = 170 N/m is pulled 17 cm from equilibrium and released. Find its maximum speed vmax.
A 40 kg mass is supported by a spring with a stiffness of 200 kN/m and a viscous damper with a damping coefficient of 280 N·s/m. Find the damping ratio ζ.
A 40 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 30 kN/m and k₂ = 25 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 56.25 m
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Answer: t = 3.081 s
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: a = 3.257 m/s²
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Down the slope is positive.
Answer: T = 30.71 N
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Answer: x = 147.2 m
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Starting and landing at the same height: t = 2vy/g.
Answer: x = 79.16 m
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Larger root: the ball is descending.
Answer: v = 1.505 m/s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: T = 32.83 N
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Towards the centre (up) is positive.
Answer: v = 7.535 m/s
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: v = 1.981 m/s
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
Answer: v′1 = −3 m/s
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Solving momentum and energy conservation together gives these two results.
The minus sign means the first glider bounces back.
Answer: v1 = 23.41 m/s
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Answer: α = 6.398 rad/s²
Rotation about a fixed axis: the sum of the moments about the axis equals I·α (ΣM = I·α), just as ΣF = m·a for translation. In energy form, a torque T turning through an angle θ (in radians) does work T·θ, which changes the kinetic energy ½·I·ω²; gravity adds or removes m·g·Δh of the centre of mass. Power is P = T·ω.
At the moment of release ω = 0, so there is no centripetal acceleration and the centre accelerates straight down.
Answer: μs = 0.1865
Rolling and angular momentum: a body rolling without slipping has v = ω·r, so its kinetic energy is ½·m·v² + ½·I·ω² = ½·m·v²·(1 + I/(m·r²)). Down a slope that gives a = g·sin θ/(1 + I/(m·r²)). When no external moment acts about an axis, the angular momentum about that axis stays constant, I₁·ω₁ = I₂·ω₂, even if kinetic energy is lost.
Answer: P = 2.545 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Answer: Tout = 80.24 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Answer: g = 3.699 m/s²
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: vmax = 1.53 m/s
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: ζ = 0.0495
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
ζ < 1, so the system is underdamped and oscillates as it decays.
Answer: fn = 3.829 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
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